I am stuck on a problem that I want to solve for my research purposes. Here it is:
Let $p \in (0,1)$ and define $f(x) = x\log\frac{x}{p} + (1-x)\log\frac{1-x}{1-p}$ for $x \in [0,1]$. It is easy to see that $f$ is strictly convex on $[0,1]$ with a unique minimum at $x=p$.
Now, suppose that you choose any arbitrary $x_1, x_2 \in [0,1]$ with $x_1 \neq x_2$. I want to show that: $$f(x_1) + f(x_2) > 2~f\left(\sqrt{\frac{x_1^2 + x_2^2}{2}}\right)~.$$
Here is what I have tried. I substituted $x_1^2 + x_2^2 = r^2$ with $r\geq 0$ and took $\theta= \cos^{-1}\left(\frac{x_1}{r}\right)$. Then, in order to solve my original problem, it suffices to show that for every fixed $r > 0$ $f(r\cos\theta) + f(r\sin\theta)$ is minimzed uniquely at $\theta = \pi/4$ over the range $\theta \in \left(\cos^{-1}\left(\frac{1}{r}\right)~,~\sin^{-1}\left(\frac{1}{r}\right)\right)$. If I take $g(\theta) = f(r\cos\theta) + f(r\sin\theta)$, then I can show that $g'(\pi/4) = 0$. Hence, the problem reduces to showing that $g$ is strictly convex over the range $\left(\cos^{-1}\left(\frac{1}{r}\right)~,~\sin^{-1}\left(\frac{1}{r}\right)\right)$. This is what I cannot show! The second derivative of $g$ becomes too ugly to handle. However, I plotted the function in MATLAB using many values of $r$, and all the plots are showing that $g$ is strictly convex.
Anyway, this is what I have tried. You can either proceed by my method, or take a different route, but I am desparate for a solution. Any solution will be highly appreciated.